What is the sum of the geometric series in which a1 = −4, r = 2, and an = −128?

Hint: r ≠ 1, where a1 is the first term and r is the common ratio.
Sn = 1,020
Sn = −252
Sn = 43,690
Sn = −1,020

Given:
The geometric series is defined by
a = a₁ = -4
r = 2

The n-th term is
[tex]a_{n} = ar^{n-1}[/tex]
Because the n-th term is -128, therefore
-4(2ⁿ⁻¹) = -128
2ⁿ⁻¹ = 32 = 2⁵
n-1 = 5
n = 6

The sum of the first n terms is
[tex]S_{n} = \frac{a(1-r^{n})}{1-r} [/tex]

Therefore
[tex]S_{n} = \frac{-4(1-2^{6})}{1-2} =4(1-64) = -252[/tex]

Answer: -252

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What is the sum of the geometric series in which a1 = −4, r = 2, and an = −128? Hint: r ≠ 1, where a1 is the first term and r is the common ratio. Sn = 1,020 Sn = −252 Sn = 43,690 Sn = −1,020

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What is the sum of the geometric series in which a1 = −4, r = 2, and an = −128?

Hint: r ≠ 1, where a1 is the first term and r is the common ratio.
Sn = 1,020
Sn = −252
Sn = 43,690
Sn = −1,020